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Question: The first and last term of an AP are 8 and 65 respectively. If the sum of all its terms in 730, find...

The first and last term of an AP are 8 and 65 respectively. If the sum of all its terms in 730, find its common difference.

Explanation

Solution

The product of the members of a finite arithmetic progression with an initial element a1,a_{1}, common differences d, and nn elements in total is determined in a closed expression
This is a generalization from the fact that the product of the progression 1×2××n1 \times 2 \times \cdots \times n is given by the factorial n ! and
that the product m×(m+1)×(m+2)××(n2)×(n1)×nm \times(m+1) \times(m+2) \times \cdots \times(n-2) \times(n-1) \times n for positive integers mm and nn is given byn!(m1)!\dfrac{n !}{(m-1) !}

Complete step-by-step answer:
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence
of numbers such that the difference between the consecutive terms is constant. For instance,
the sequence 5,7,9,11,13, 15,…is an arithmetic progression with a common difference of 2 .
If the initial term of an arithmetic progression is a1a_{1} and the common difference of successive
members are d, then the nth term of the sequence (an)\left(a_{n}\right) is given by:
an=a1+(n1)da_{n}=a_{1}+(n-1) d and in general,an=am+(nm)da_{n}=a_{m}+(n-m) d.
Given 730=n2(a+l)730=\dfrac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})
l=65\mathrm{l}=65
a=8\mathrm{a}=8
\Rightarrow 730=n2(8+65)730=\dfrac{\mathrm{n}}{2}(8+65)
\Rightarrow 730=n2(73)730=\dfrac{\mathrm{n}}{2}(73)
\Rightarrow 10=n210=\dfrac{\mathrm{n}}{2}
20=n20=\mathrm{n}
Now 65=l=a+(n1)d65=\mathrm{l}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}
\Rightarrow 65=8+(201)d65=8+(20-1) \mathrm{d}
\Rightarrow 658=19d65-8=19 \mathrm{d}
\Rightarrow 57=19d57=19 \mathrm{d}
\Rightarrow d=5719=3\mathrm{d}=\dfrac{57}{19}=3
The common difference is 3.

Note: A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series. The standard deviation of any arithmetic progression can be
calculated as σ=d(n1)(n+1)12\sigma=|d| \sqrt{\dfrac{(n-1)(n+1)}{12}}
where nn is the number of terms in the progression and dd is the common difference between terms. The formula is very similar to the standard deviation of a discrete uniform distribution.